證明題(sin.cos)

2010-06-27 7:01 am
試證:三角形ABC中
(1)cosA+cosB+cosC=1+4sinA/2sinB/2sinC/2
(2)sinA+sinB+sinC=4cosA/2cosB/2cosC/2

回答 (1)

2010-06-27 7:19 am
✔ 最佳答案
A+B+C=π
1.
cosA+cosB+cosC
=2cos[(A+B)/2]cos[(A-B)/2]+1-2[sin(C/2)]^2
=1+ 2cos[π/2-C/2)cos(A/2 - B/2)-2sin(C/2)*sin[π/2 -(A+B)/2]
=1+ 2sin(C/2)cos(A/2 - B/2)- 2sin(C/2)*cos(A/2 + B/2)
=1+ 2sin(C/2)[cos(A/2 - B/2) - cos(A/2 + B/2)]
=1+2sin(C/2)[cos(A/2)cos(B/2)+sin(A/2)sin(B/2)-cos(A/2)cos(B/2)+sin(A/2)sin(B/2)]
=1+4sin(C/2)sin(A/2)sin(B/2)

2.
sinA+sinB+sinC
=2sin[(A+B)/2]cos[(A-B)/2]+2sin(C/2)cos(C/2)
=2sin(π/2 - C/2)cos(A/2 - B/2)+2cos(C/2)sin[π/2 -(A+B)/2]
=2cos(C/2)cos(A/2 - B/2)+2cos(C/2)cos(A/2 + B/2)
=2cos(C/2)[cos(A/2 - B/2)+cos(A/2 + B/2)]
=4cos(C/2)cos(A/2)cos(B/2)


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